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Consider the following code:

Fs = 48000;
h = dsp.ParametricEQFilter('CenterFrequency', 1000, ...
                           'Bandwidth', 1400, ...
                           'PeakGaindB', 10, ...
                           'SampleRate', Fs);

fvtool(h,'Fs',Fs,'FrequencyScale','Log');

Let the two bandwidth $-3\rm dB$ points $f_1$ and $f_2$. Therefore: $f_1 = 300\ \mathrm{Hz}, \ f_2 = 1700\ \textrm{Hz}$

However, in the graph I see the $-3\ \mathrm{dB}$ point (i.e. the $+7\ \rm dB$ gain) at $487\ \rm Hz$. What am I doing wrong?

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  • $\begingroup$ Just a guess as I can't run that code, but is the graph using 10 log(|x|) rather than 20 log(|x|)? $\endgroup$ – AnonSubmitter85 Dec 18 '17 at 21:14
  • $\begingroup$ @AnonSubmitter It is using 20 log(|x|). You can also get the frequency response by: [cVal, normFreq] = h.freqz; freqaxis = normFreq/ pi * (Fs/2); dBval = 20*log10(abs(cVal)); semilogx(freqaxis, dBval); $\endgroup$ – CyDe Dec 19 '17 at 13:06
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Therefore: $f_1 = 300\ \rm Hz$, $f2 = 1700\ \rm Hz$

No. Bandwidth is relative not absolute. The center frequency is the geometric mean of the corner frequencies, not the arithmetic mean. For $1\ \rm kHz$ and $1400\ \rm Hz$ of bandwidth I would expect the corner frequencies to be $520\ \rm Hz$ to $1920\ \rm Hz$. The ratio of $1920$ to $1000$ is the same as the ratio of $1000$ to $520$.

This solves the equation $\ \displaystyle x - \frac{1}{x} = 1.4$

I also would suggest that defining bandwidth for a parametric filter using "$-3\ \rm dB$" doesn't make a lot of sense. What if the peak gain is only $2\ \rm dB$? I'm guessing that you may misinterpret the definition of bandwidth in this context. More commonly would be "frequencies where the gain is half of the peak gain".

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Apparently, after some research, I have found that the bandwidth "bandedges" is a bit of a tricky subject: (see Bandwidth Re-visited). Some references suggest indeed that the bandwidth is measured at the gain which is equal to half of the peak gain. Some others say that you should measure the bandwidth at $ +3\ \textrm{dB}$ for boost gain, $ -3\ \textrm{dB}$ for cut. (which eventually runs into the issue of where it should be measured if the gain is less than $ 3\ \textrm{dB}$.

Hilmar is right about the geometric mean of the corner frequencies. However, the cutoff points (i.e. the bandedges) do not exist at half the peak gain in MATLAB. They exist at $G^{2}_B = \frac{1+G^2}{2}$, where $G$ is the parametric equalizer linear gain, and $G_B$ is the linear bandwidth gain as described here.

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    $\begingroup$ The linked paper in the answer above is an old one of mine. if you view it, do not use the pdf viewer in Chrome because it's shit in rendering the equations (which were set using Equation Editor back in the previous century). Download and use Acrobat Reader to view the paper. $\endgroup$ – robert bristow-johnson Dec 19 '17 at 17:30

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